The coordinates of the focus of the parabola | vedclass

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Question

(D) Given the equation of the parabola: $5x^2 = -12y$. Dividing by $5$,we get $x^2 = -\frac{12}{5}y$. Comparing this with the standard form $x^2 = 4ay

Vedclass · Accepted Answer

$\left(0, -\frac{3}{5}\right)$

Vedclass · Answer

(D) Given the equation of the parabola: $5x^2 = -12y$. Dividing by $5$,we get $x^2 = -\frac{12}{5}y$. Comparing this with the standard form $x^2 = 4ay$,we have $4a = -\frac{12}{5}$. Solving for $a$,we get $a = -\frac{12}{5 	imes 4} = -\frac{3}{5}$. The focus of a parabola of the form $x^2 = 4ay$ is given by $(0, a)$. Substituting the value of $a$,the focus is $\left(0, -\frac{3}{5}ight)$. Therefore,option $D$ is correct.

The coordinates of the focus of the parabola $5x^2 = -12y$ are

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